JEE Questions for Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 10 - MCQExams.com

Three, These are : CH₃OCH₂CH₂CH₃,
CH₃-O-CH(CH₃)₂ and CH₃CH₂OCH₂CH₃.

Structures (a), (c) and (d) have the same M.F. (C₆H₁₂O) as that of CH₃COCH₂CH₂CH₂CH₃ and hence are all isomers.The M.F of structure (b) is C₆H_10O and hence it is not an isomer.

1 and 2 represent staggered conformation

meso-Tartaric acid optically inactive due to molecular symmetry because of the presence of a plane of symmetry in its molecule.

Since during the reaction, a chiral carbon is created and further since the CN - ion can attack the planar aldehyde group both from the top and the bottom face of the aldehyde group with equal ease, therefore, a 50:50 mixture of the two enantiomers, i.e. a racemic mixture is obtained.

trans-2-Butene has zero dipole moment.

Three. These are: d-tartaric acid, l-tartaric acid and meso-tartaric acid.

The two compounds are non-superimposable mirror images of each other and hence are enantiomers.

Conformers

The two compounds are identical since they have a plane of symmetry.

Boat conformation of cyclohexane has two types of destabilizing intereactions, i.e, eclipsing and flagpole-flagpole. Out of these eclipsing interactions are the most destabilizing.

Only DCH₂CH₂CH₂Cl does not contain a chiral carbon and hence it is not chiral

As the size of the ring increases, the internal angle increases accordingly. As a result, the derivation from the tetrahedral angle and hence the angle strain increases as the size of the ring increases. Thus planar cyclodecane is expected to have the maximum angle strain.

The configuration of the penultimate carbon of any optically active compound is always correlated with D (+)-glyceraldehyde.
Therefore, option (c) is correct.

The two compounds are non-superimposable mirror images of each other and hence are enantiomers.

The given compound contains a double bond (with different substituents on each carbon) as well as a chiral carbon, therefore, it shows both geometrical as well as optical isomerism.

(R) - and (S) - enantiomers differ only in their behaviour towards plane polarized light.

Restricted rotation around C=C bond

Three: CH₃OCH₂CH₂CH₃,CH₃CH₂OCH₂CH₃CH₃OCH (CH₃)₂

CH₃ – C ≡ C – CH₃ being linear as well as symmetrical, has lowest dipole moment.

II and IV and cis-trans-isomers

I and II contain chiral C-atoms and hence are chiral compounds

All the three exhibit conformational isomerism due to rotation about C - O bond in CH₃OH, O - O bond in H₂O₂ and N - N bond in NH₂ – NH₂.

Disubstituted cyclic compounds and disubstituted alkenes show geometrical isomerism. Therefore, option (d) correct.

Enantiomers have same m.p./b.p and refractive indices but rotate plane polarized light in opposite directions but to the same extent.

(+) and(-) Lactic acid on treatment with an optically active base such as (+) or (-) brucine to form diastereomers which have different m.p/b.p and solubilities and hence can be separated

Compounds possessing one chiral carbon atom are always optically active.

Glycine is optically inactive but alanine is optically active, Therefore, the dipeptide glycylalanine is also optically active.